![EE 201 superposition – 15
Example 5
Find vR2.
This looks nasty – there are 6 nodes with a super node or 4 meshes with
a super mesh. Either NV or MC will be tough. We might be able to do a
bunch of source transformations, but since there are only three sources,
let’s try superposition.
100 V
50 Ω
75 Ω
30 Ω
50 Ω
75 Ω
90 Ω 75 V
0.6 A
Start with VS1. Deactivate IS and
VS2. Shorting VS2 puts R6 in
parallel with R5. Use a voltage
divider between R1 and all the
rest. Need the equivalent
resistance of R2 in parallel with
all the other stuff.
v′
R2 =
R26
R26 + R1
VS1 =
50 Ω
50 Ω + 50 Ω
(100 V) = 50 V
R26 = R2∥ (R3 + R4 + R5∥R6) = (75 Ω) ∥ [30 Ω + 90 Ω + (50 Ω) ∥ (75 Ω)] = 50 Ω
+
–
VS1
R1
R3
R2
R4
R5
+
–
IS
R6
VS2
–
+
vR2
+
–
VS1
R1
R3
R2
R4
R5
R6
–
+
v′
R2
+
–
VS1
R1
R26
–
+
v′
R2](https://image.slidesharecdn.com/superposition-240623194032-b309d50c/85/superposition-principle-notes-pdf-15-320.jpg)
![EE 201 superposition – 16
Now VS2. Go back to the
original circuit and deactivate
IS and VS1. Shorting VS1 puts
R1 in parallel with R2.
Use voltage dividers twice: first to find the voltage across R5, then again
to find voltage across R2. Need the equivalent resistance of R5 and all of
the stuff in parallel with it, seen looking back from VS2.
50 Ω
75 Ω
30 Ω
50 Ω
75 Ω
90 Ω
75 V
+
– VS2
R1
R3
R2
R4
R5
R6
–
+
v′′
R2
–
+
v′′
R5
R15 = R5∥ (R3 + R4 + R1∥R2)
+
– VS2
R15
R6
–
+
v′′
R5
= (50 Ω) ∥ [30 Ω + 90 Ω + (50 Ω) ∥ (75 Ω)] = 37.5 Ω
v′′
R5 =
R15
R15 + R6
VS2 =
37.5 Ω
37.5 Ω + 75 Ω
(75 V) = 25 V
v′′
R2 =
R1∥R2
R1∥R2 + R3 + R4
v′′
R5
=
30 Ω
30 Ω + 30 Ω + 90 Ω
(25 V) = 5 V
R1
R3
R2
R4
–
+
v′′
R2
–
+
v′′
R5 = 25 V
Finally,
R1∥R2 = (50 Ω) ∥ (75 Ω) = 30 Ω](https://image.slidesharecdn.com/superposition-240623194032-b309d50c/85/superposition-principle-notes-pdf-16-320.jpg)
superposition principle notes .pdf
- 1. EE 201 superposition – 1 Superposition • Equivalent resistance • Voltage / current dividers • Source transformations • Node voltages • Mesh currents • Superposition In a circuit having more than one independent source, we can consider the effects of the sources one at a time.
- 2. EE 201 superposition – 2 2va + 6vb = 2 V 8va 2vb = 36 V Solving gives: va = 5 V, vb = 2 V 8va 2vb = 0 2va + 6vb = 2 V Solving gives: and 2va + 6vb = 0 8va 2vb = 36 V A math problem: Mathematically, we can solve the simultaneous equations a piece at a time. If these equations are from a circuit, this math implies that we might be able to solve the circuit a piece at time. v′ a = 0.09091V v′ b = 0.36364V Solving gives: and v′′ a = 4.90909V v′′ b = 1.63636V v′ a + v′′ a = 5.0V = va v′ b + v′′ b = 2.0V = vb
- 3. EE 201 superposition – 3 The superposition method In a circuit having more than one independent source, we can consider the effects of the sources one at a time. If a circuit has n independent sources, then we will have to solve n separate circuits. It this easier? Perhaps. The resulting “partial” circuits will have one source and some resistors. We might be able to solve the partial circuits using the short-cut methods we saw earlier – each partial circuit may be very easy. As we consider the effect of each source by itself, we must “turn off” (deactivate) all of the other sources. Deactivation means setting the values to zero. same as same as short circuit open circuit Replace voltage sources with shorts. Replace current sources with opens. IS 0 A + – VS 0 V
- 4. EE 201 superposition – 4 Example of the superposition method Consider the familiar two- source, two-resistor circuit one more time. Let’s try superposition to find vR2. 1. Deactivate (turn off) one of the sources — the order doesn’t matter. Let’s deactivate the current source first — set the value of IS to zero, which has the same effect as replacing IS with an open circuit. The result is a simple voltage divider, and we can easily calculate the partial result due to the voltage source. 10 V 1 A 10 ! 5 ! + – VS R1 R2 – + v′ R2 + – VS IS R1 R2 – + vR2 v′ R2 = R2 R1 + R2 VS = 5 Ω 5 Ω + 10 Ω (10 V) = 3.33 V
- 5. EE 201 superposition – 5 2. Go back to the original circuit and turn off the other source – set VS to zero, which is the same as replacing it with a short circuit. Shorting VS causes R1 to be in parallel with R2. 3. The complete answer is the sum (superposition) of the two partial answers. IS R1 R2 – + v′′ R2 v′′ R2 = IS (R1∥R2) = (1 A) (5 Ω∥10 Ω) = 3.33 V vR2 = v′ R2 + v′′ R2 = 3.33 V + 3.33 V = 6.66 V Of course, this is identical to the values obtained by all of the other methods that were used to analyze this same circuit. 1 A 10 ! 5 !
- 6. EE 201 superposition – 6 Summary of the superposition method 1. Identify all of the independent sources in the circuit. 2. Choose one source that will remain active. Deactivate all of the others. (Remove current sources, leaving open circuits. Replace voltage sources with short circuits.) 3. Using whatever techniques are appropriate, solve for the desired quantity (current or voltage) in the circuit. This will be a “partial” result, due only to the one active source in the circuit. 4. Return to the original circuit. Choose a different source to remain active and deactivate all of the others. 5. Solve again for the desired quantity, which will be a second partial result. 6. Continue in this manner, working sequentially through each of the sources in the circuit, finding a partial result for each. 7. Add together all of the partial results to obtain the total result corresponding to all of the sources working simultaneously in the circuit.
- 7. EE 201 superposition – 7 Cautions in using superposition 1. Superposition only works with linear circuits. (Linear circuits contain only sources, resistors, capacitors, inductors, linear amplifiers, etc.) Most electronic devices (diodes and transistors) are non-linear, so superposition will not be applicable. 2. Because the method relies on linearity, you cannot add powers directly using the superposition method. (Power goes as v2 or i2 – it is not linear.) Use superposition to find the total current or voltage and then calculate power from that result. 3. When finding the partial solutions, be sure to maintain the same voltage polarity and current direction in each case. For example, one source may induce the current in a particular resistor to flow in one direction while another source causes to a current flowing in the opposite direction. You must keep the proper signs when adding to the partial results to obtain the correct total result.
- 8. EE 201 superposition – 8 + – VS1 VS2 R1 R3 – + vR1 + – R2 14 V 6 k! Example 2 9 k! 9 k! 7 V For the circuit shown, use superposition to find the value of vR1. 1. Deactivate VS2. The result is a voltage divider between R1 and the parallel combination of R2 and R3. v′ R1 = R1 R1 + R2∥R3 VS1 = 6 kΩ 6 kΩ + (9 kΩ∥9 kΩ) (14 V) = 8 V + – VS1 R1 R3 – + R2 v′ R1
- 9. EE 201 superposition – 9 VS2 R1 R3 – + + – R2 v′′ R1 2. Now deactivate VS1. Again, the result is a simple voltage divider, but now R1 is in parallel with R3. More importantly, note the polarity of — it will be negative for this partial solution. v′′ R1 6 k! 9 k! 9 k! 7 V v′′ R1 = − R1∥R3 R1∥R3 + R2 VS2 = − (6 kΩ∥9 kΩ) (6 kΩ∥9 kΩ) + 9 kΩ (7 V) = − 2 V 3. The complete solution is the sum of the two partial answers. vR1 = v′ R1 + v′′ R1 = 8 V − 2 V = 6 V
- 10. EE 201 superposition – 10 50 V Example 3 For the circuit shown, use superposition to find the power being dissipated in R3. + – VS1 VS2 R1 R3 + – R2 R4 R5 1. In using superposition, we cannot find “partial powers” — we need to find either total voltage or total current and then calculate power. For this problem, we will choose to find vR3. Start by deactivating VS2. We note that the series combination of R2 and R5 is in parallel with R3. – + + – VS1 R1 R3 R2 R4 R5 v′ R3 – + + – VS1 R1 R235 R4 v′ R3 R235 = R3∥ (R2 + R5) = (20 Ω) ∥ (35 Ω + 65 Ω) = 16.67 Ω v′ R3 = R235 R235 + R1 + R4 VS1 = 16.67 Ω 16.67 Ω + 10 Ω + 40 Ω (50 V) = 12.5 V 100 V 10 Ω 35 Ω 40 Ω 65 Ω 20 Ω
- 11. EE 201 superposition – 11 2. Go back to the original circuit and deactivate VS1. Note that with VS1 shorted, the series combination of R1 and R4 is in parallel with R3. v′′ R3 = R134 R134 + R2 + R5 VS2 = 14.29 Ω 14.29 Ω + 35 Ω + 65 Ω (100 V) = 12.5 V VS2 R1 R3 + – R2 R4 R5 – + v′′ R3 VS2 R134 + – R2 R5 – + v′′ R3 R134 = R3∥ (R1 + R4) = (20 Ω) ∥ (10 Ω + 40 Ω) = 14.29 Ω 3. The complete solution is the sum of the two partial answers. vR3 = v′ R3 + v′′ R3 = 12.5 V + 12.5 V = 25 V PR3 = v2 R3 R3 = (25 V) 2 20 Ω = 31.25 W 100 V 10 Ω 35 Ω 40 Ω 65 Ω 20 Ω And finally,
- 12. EE 201 superposition – 12 Example 4 In the circuit, find iR3. With three sources, there will be three partial solutions. 1. Start with VS. Deactivate the two current sources. Find the partial current, using whatever short-cut methods you like. One approach: use a source transformation and then a current divider. 10 V 1 kΩ 3 kΩ 15 mA 7.5 mA 1 kΩ 2 kΩ + – VS IS2 IS1 R1 R2 R3 R4 iR3 + – VS R1 R2 R3 R4 i′ R3 I′ S = VS R1 = 10 V 1 kΩ = 10 mA R1 R2 R3 R4 i′ R3 I′ S i′ R3 = 1 R3 + R4 1 R3 + R4 + 1 R1 + 1 R2 I′ S = 1 3 kΩ 1 3 kΩ + 1 1 kΩ + 1 3 kΩ (10 mA) = 2 mA
- 13. EE 201 superposition – 13 2. Go back to the original circuit and de- activate VS and IS2, keeping IS1. Note that shorting VS places R1 is in parallel with R2. Examining the partial circuit, we see that it is essentially a current divider. The source current splits between two paths – one with R3 and another that is a combination of R1, R2, and R4. 1 kΩ 3 kΩ 15 mA 1 kΩ 2 kΩ IS1 R1 R2 R3 R4 i′′ R3 R124 R124 = R1∥R2 + R4 = 1 kΩ∥3 kΩ + 2 kΩ = 2.75 kΩ i′′ R3 = 1 R3 1 R3 + 1 R124 IS1 = 1 1 kΩ 1 1 kΩ + 1 2.75 kΩ (15 mA) = 11 mA v′ R3 IS1 R1 R2 R3 R4 i′′ R3
- 14. EE 201 superposition – Again, we can find the current with a current divider. The source current splits between two paths – the branch with R4 and another that is a combination of R1, R2, and R3. Also, note carefully the direction – will be negative! 14 3. To obtain the final partial circuit, go back to the original circuit and de- activate VS and IS1, keeping IS2. As in the prior case, R1 is in parallel with R2. 4. To complete the calculation, add the three partial results to get the total. 1 kΩ 3 kΩ 7.5 mA 1 kΩ 2 kΩ R123 = R1∥R2 + R3 = 1 kΩ∥3 kΩ + 1 kΩ = 1.75 kΩ i′′′ R3 i′′′ R3 = − 1 R123 1 R123 + 1 R4 IS2 = − 1 1.75 kΩ 1 1.75 kΩ + 1 2 kΩ (7.5 mA) = − 4 mA iR3 = i′ R3 + i′′ R3 + i′′′ R3 = 2 mA + 11 mA − 4 mA = 9 mA IS2 R1 R2 R3 R4 i′′′ R3 IS2 R1 R2 R3 R4 i′′′ R3 R123
- 15. EE 201 superposition – 15 Example 5 Find vR2. This looks nasty – there are 6 nodes with a super node or 4 meshes with a super mesh. Either NV or MC will be tough. We might be able to do a bunch of source transformations, but since there are only three sources, let’s try superposition. 100 V 50 Ω 75 Ω 30 Ω 50 Ω 75 Ω 90 Ω 75 V 0.6 A Start with VS1. Deactivate IS and VS2. Shorting VS2 puts R6 in parallel with R5. Use a voltage divider between R1 and all the rest. Need the equivalent resistance of R2 in parallel with all the other stuff. v′ R2 = R26 R26 + R1 VS1 = 50 Ω 50 Ω + 50 Ω (100 V) = 50 V R26 = R2∥ (R3 + R4 + R5∥R6) = (75 Ω) ∥ [30 Ω + 90 Ω + (50 Ω) ∥ (75 Ω)] = 50 Ω + – VS1 R1 R3 R2 R4 R5 + – IS R6 VS2 – + vR2 + – VS1 R1 R3 R2 R4 R5 R6 – + v′ R2 + – VS1 R1 R26 – + v′ R2
- 16. EE 201 superposition – 16 Now VS2. Go back to the original circuit and deactivate IS and VS1. Shorting VS1 puts R1 in parallel with R2. Use voltage dividers twice: first to find the voltage across R5, then again to find voltage across R2. Need the equivalent resistance of R5 and all of the stuff in parallel with it, seen looking back from VS2. 50 Ω 75 Ω 30 Ω 50 Ω 75 Ω 90 Ω 75 V + – VS2 R1 R3 R2 R4 R5 R6 – + v′′ R2 – + v′′ R5 R15 = R5∥ (R3 + R4 + R1∥R2) + – VS2 R15 R6 – + v′′ R5 = (50 Ω) ∥ [30 Ω + 90 Ω + (50 Ω) ∥ (75 Ω)] = 37.5 Ω v′′ R5 = R15 R15 + R6 VS2 = 37.5 Ω 37.5 Ω + 75 Ω (75 V) = 25 V v′′ R2 = R1∥R2 R1∥R2 + R3 + R4 v′′ R5 = 30 Ω 30 Ω + 30 Ω + 90 Ω (25 V) = 5 V R1 R3 R2 R4 – + v′′ R2 – + v′′ R5 = 25 V Finally, R1∥R2 = (50 Ω) ∥ (75 Ω) = 30 Ω
- 17. EE 201 superposition – 17 Finally, IS. Deactivate VS1 and VS2 in the original circuit. The source current divides to the left and right according to the equivalent resistances on either side. First find the equivalent resistances left and right: To finish, add the three partial results to get the total. R1 R3 R2 R4 R5 IS R6 – + v′′′ R2 iL iR Use a current divider to determine the amount flowing to the left: RL = R3 + R1∥R2 RR = R4 + R5∥R6 = 30 Ω + (50 Ω) ∥ (75 Ω) = 60 Ω 50 Ω 75 Ω 30 Ω 50 Ω 75 Ω 90 Ω 0.6 A = 90 Ω + (50 Ω) ∥ (75 Ω) = 120 Ω iL = 1 RL 1 RL + 1 RR IS = 1 60 Ω 1 60 Ω + 1 120 Ω (0.6 A) = 0.4 A R12 R3 – + v′′′ R2 iL v′′′ R2 = iL (R1∥R2) = (0.4 A) (30 Ω) = 18 V vR2 = v′ R2 + v′′ R2 + v′′′ R2 = 50 V + 5 V + 18 V = 73 V